What is the tangent of f(x)=((x+1)/(x-1))^3,\at x=2 ?

The tangent of f(x)=((x+1)/(x-1))^3,\at x=2 is y=-54x+135

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tangent of f(x)=((x+1)/(x-1))^3,\at x=2

tangent of

f (x) = (\frac{x + 1}{x - 1})^{3}, at x = 2

y = - 54 x + 135

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

f (x) = (\frac{x + 1}{x - 1})^{3} : \frac{df ( x )}{d x} = - \frac{6 ( x + 1 ) ^{2}}{( x - 1 ) ^{4}}

Compute the slope of

f (x) = (\frac{x + 1}{x - 1})^{3}

x = 2 : m = - 54

Find the line with slope m=

- 54

and passing through

(2, 27) : y = - 54 x + 135

y = - 54 x + 135

\int 6 x^{2} e^{x^{3}} d x

\int \frac{1}{3 - 2 x - x ^{2}} d x

\frac{\partial}{\partial x} (x y e^{yz})

\int_{- 1}^{5} ∣ 2 x - 3 ∣ d x

\frac{d}{d t} (\frac{1 - 2 t}{1 + 2 t})

What is the tangent of f(x)=((x+1)/(x-1))^3,\at x=2 ?
The tangent of f(x)=((x+1)/(x-1))^3,\at x=2 is y=-54x+135